Question

If every solution of the equation $3{\cos }^{2}x-\cos x-1=0$ is a solution of the equation $a{\cos }^{2}2x+b\cos 2x-1=0$. Then find the value of $(a+b)$.

Answer 1

Answer: 13

$1^{\text{st }}$ equation gives,
$\frac{3}{2}(1+\cos 2x)-1=\cos x\Rightarrow 3(1+\cos 2x)-2=2\cos x$
$3\cos 2x+1=2\cos x$
$(3\cos 2x+1{)}^2=4{\cos }^{2}x=2(1+\cos 2x)$
$9{\cos }^{2}2x+4\cos 2x-1=0$ ....(1)
$\text{ comparing with }$
$a{\cos }^{2}2x+b\cos 2x-1=0$ ...(2)
$\frac{a}{9}=\frac{b}{4}=1\Rightarrow b=4\text{ and }a=9\Rightarrow a+b=13$